HDU - 5667 Sequence(矩阵快速幂+费马小定理)

写这题的时候压根不知道费马小定理是啥

HDU - 5667 

Problem Description

     Holion August will eat every thing he has found.

     Now there are many foods,but he does not want to eat all of them at once,so he find a sequence.

fn=⎧⎩⎨⎪⎪1,ab,abfcn−1fn−2,n=1n=2otherwise

     He gives you 5 numbers n,a,b,c,p,and he will eat  fn  foods.But there are only p foods,so you should tell him  fn  mod p.

 

Input

     The first line has a number,T,means testcase.

     Each testcase has 5 numbers,including n,a,b,c,p in a line.

    1≤T≤10,1≤n≤1018,1≤a,b,c≤109 , p  is a prime number,and  p≤109+7 .

 

Output

     Output one number for each case,which is  fn  mod p.

 

Sample Input

  
  

   
   1
5 3 3 3 233
  
  

 

Sample Output

  
  

   
   190
  
  
  
  


  
  

题意：很简单不说了

思路：推出a的系数，用矩阵快速幂求a的系数（系数还不是我推的，队友推的），但是这个时候发现不能直接用系数去modp，所以需要get新知识，早上写题真是WA到傻才明白我可能漏了些知识。这里的系数需要mod（p-1），因为首先p是一个质数，所以如果a不是p的倍数的话，肯定与p满足互质这个条件，所以满足费马小定理的式子，所以每一个a^(p-1)%(p)肯定是1的，所以a的系数里每p-1个对modp的贡献都是1，所以对系数需要mod（p-1）。

代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
#define ll long long

ll Mod,c;
struct node
{
    ll mat[3][3];
    node()
    {
        for(int i = 0; i < 3; i++)
        for(int j = 0; j < 3; j++)
            mat[i][j] = 0;
    }
};
node operator*(node a,node b)
{
    node temp;
    for(int i = 0; i < 3; i++)
    for(int j = 0; j < 3; j++)
    for(int k = 0; k < 3; k++)
        temp.mat[i][j] = (temp.mat[i][j] + (a.mat[i][k] * b.mat[k][j])%(Mod-1))%(Mod-1);
    return temp;
}
ll qpow(ll a,ll b)
{
    ll ans = 1;
    while(b)
    {
        if(b&1) ans = (ans*a)%Mod;
        a = (a*a)%Mod;
        b = b >> 1;
    }
    return ans;
}
ll mpow(ll n)
{
    if(n == 1) return 0;
    if(n == 2) return 1;
    n = n-2;
    node e,ans;
    ll temp[3] = {1,0,1};
    e.mat[0][0] = c, e.mat[0][1] = 1, e.mat[0][2] = 1;
    e.mat[1][0] = 1, e.mat[2][2] = 1;
    for(int i = 0; i < 3; i++) ans.mat[i][i] = 1;
    while(n)
    {
        if(n&1) ans = ans*e;
        e = e*e;
        n = n >> 1;
    }
    ll ret = 0;
    for(int i = 0; i < 3; i++)
        ret = (ret + (ans.mat[0][i]*(temp[i]%(Mod-1)))%(Mod-1))%(Mod-1);
    return ret;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll a,b,n;
        cin >> n >> a >> b >> c >> Mod;
        if(a%Mod == 0)
        {
            cout << 0 << endl;
            continue;
        }
        ll temp = b*mpow(n)%(Mod-1);
        ll ans = qpow(a,temp);
        cout << ans << endl;
    }
	return 0;
}